tính hợp lý :
a, \(2008.\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009.\left(\dfrac{1}{2007}-2\right)\)
b, \(\dfrac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3.4^5}\)
Tính hợp lý:
a, \(2008\cdot\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009\cdot\left(\dfrac{1}{2007}-2\right)\)
b,\(\dfrac{5^5\cdot20^3-5^4\cdot20^3+5^7\cdot4^5}{\left(20+5\right)^3\cdot4^5}\)
1. 2008.\(\left(\dfrac{1}{2007}-\dfrac{2009}{1004}\right)-2009\left(\dfrac{1}{2007}-2\right)\)
=\(\left(2008.\dfrac{1}{2007}-2008.\dfrac{2009}{1004}\right)-\left(2009.\dfrac{1}{2007}-2009.2\right)\)
=\(\left(\dfrac{2008}{2007}-2.2009\right)-\left(\dfrac{2009}{2007}-2.2009\right)\)
=\(\left(\dfrac{2008}{2007}-4018\right)-\left(\dfrac{2009}{2007}-4018\right)\)
=\(\dfrac{2008}{2007}-4018-\dfrac{2009}{2007}+4018\)
=\(\left(\dfrac{2008}{2007}-\dfrac{2009}{2007}\right)+\left[\left(-4018\right)+4018\right]\)
=\(\dfrac{1}{2007}.\left(2008-2009\right)+0\)
=\(\dfrac{1}{2007}.\left(-1\right)+0\)
=\(\dfrac{-1}{2007}\)
2.\(\dfrac{5^5.20^3-5^4.20^3+5^7.4^5}{\left(20+5\right)^3+4^5}\)
=\(\dfrac{5^5.\left(2^2.5\right)^3-5^4.\left(2^2.5\right)^3+5^7.\left(2^2\right)^5}{\left[\left(2^2.5\right)+5\right]^3+\left(2^2\right)^5}\)
=\(\dfrac{5^5.2^6.5^3-5^4.2^6.5^3+5^7.2^{10}}{2^6.5^3+5^3+2^{10}}\)
=\(\dfrac{5^9.2^6-5^7.2^6+5^7.2^{10}}{5^3.\left(2^6+1\right)+2^{10}}\)
=\(\dfrac{5^7.2^6\left(5^2-1-2^4\right)}{5^3\left(2^6+1\right)+2^{10}}\)
bí rồi
Bài 1:
a. \(\left(5^{2010}+5^{2012}+5^{2014}\right):\left(5^{2011}+5^{2009}+5^{2007}\right)\)
b. \(\left(-\dfrac{7}{45}\right)-\left(-\dfrac{1}{4}\right)-\left(-\dfrac{3}{5}\right)+\dfrac{1}{12}+\dfrac{2}{3}+\dfrac{1}{39}-\left(-\dfrac{5}{9}\right)\)
\(a)\left(5^{2010}+5^{2012}+5^{2014}\right):\left(5^{2011}+5^{2009}+5^{2007}\right)\)
\(=\dfrac{5^{2007}\left(5^3+5^5+5^7\right)}{5^{2007}\left(5^4+5^2+1\right)}=\dfrac{5^3+5^5+5^7}{5^4+5^2+1}\)
\(=\dfrac{125+3125+78125}{625+25+1}=\dfrac{81375}{651}=125\)
\(b)-\dfrac{7}{45}+\dfrac{1}{4}+\dfrac{3}{5}+\dfrac{1}{12}+\dfrac{2}{3}+\dfrac{1}{39}+\dfrac{5}{9}\)
\(=\dfrac{-7.52+1.585+3.468+1.195+2.780+1.60-5.260}{2340}\)
\(=\dfrac{-364+585+1404+195+1560+60-1300}{2340}\)
\(=\dfrac{2140}{2340}=\dfrac{107}{117}\)
a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
a)
PT \(\Leftrightarrow \frac{4x+2}{12}-\frac{3x-6}{12}=\frac{12-8x}{12}-\frac{12x}{12}\)
\(\Leftrightarrow 4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow 21x=4\Leftrightarrow x=\frac{4}{21}\)
b)
PT \(\Leftrightarrow \frac{30x+15}{20}-\frac{100}{20}-\frac{6x+4}{20}=\frac{24x-12}{20}\)
\(\Leftrightarrow 30x+15-100-6x-4=24x-12\Leftrightarrow -89=-12\) (vô lý)
Vậy pt vô nghiệm.
Tìm giá trị lớn nhất , giá trị nhỏ nhất của biểu thức :
a)\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\)
b)B=\(\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\)
c)C=\(-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\)
Ai lm đc câu nào thì giúp mk với , cảm ơn !!
\(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\\ A_{min}=\dfrac{1}{9}\Leftrightarrow x=\dfrac{3}{5}\\ B=\dfrac{2009}{2008}-\left|x-\dfrac{3}{5}\right|\le\dfrac{2009}{2008}\\ B_{max}=\dfrac{2009}{2008}\Leftrightarrow x=\dfrac{3}{5}\\ C=-2\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\le1\dfrac{2}{3}\\ C_{max}=1\dfrac{2}{3}\Leftrightarrow\dfrac{1}{3}x=-4\Leftrightarrow x=-12\)
a: \(A=\left|\dfrac{3}{5}-x\right|+\dfrac{1}{9}\ge\dfrac{1}{9}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{5}\)
Bài 1: Giải phương trình
\(a,\dfrac{x+1}{2009}+\dfrac{x+3}{2007}=\dfrac{x+5}{2005}+\dfrac{x+7}{1993}\)
\(b,\left(x+2\right)^4+\left(x+4\right)^4=14\)
\(c,\left(x-3\right)\left(x-2\right)x+1=60\)
d, \(2x^4+3x^3-x^2+3x+2=0\)
Với \(x=0\) không phải nghiệm
Với \(x\ne0\) chia 2 vế cho \(x^2\), pt tương đương:
\(2x^2+3x-1+\dfrac{3}{x}+\dfrac{2}{x^2}=0\)
\(\Leftrightarrow2\left(x+\dfrac{1}{x}\right)^2+3\left(x+\dfrac{1}{x}\right)-5=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=1\\x+\dfrac{1}{x}=-\dfrac{5}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\2x^2+5x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô-nghiệm\right)\\\left(x+2\right)\left(2x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Câu a chắc là đề sai, vì nghiệm vô cùng xấu, tử số của phân thức cuối cùng là \(x+17\) mới hợp lý
b.
Đặt \(x+3=t\)
\(\Rightarrow\left(t+1\right)^4+\left(t-1\right)^4=14\)
\(\Leftrightarrow t^4+6t^2-6=0\) (đến đây đoán rằng bạn tiếp tục ghi sai đề, nhưng thôi cứ giải tiếp)
\(\Rightarrow\left[{}\begin{matrix}t^2=-3+\sqrt{15}\\t^2=-3-\sqrt{15}\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow t=\pm\sqrt{-3+\sqrt{15}}\Rightarrow x=-3\pm\sqrt{-3+\sqrt{15}}\)
Câu c chắc cũng sai đề, vì lên lớp 8 rồi không ai cho đề kiểu này cả, người ta sẽ rút gọn luôn số 1 bên trái và 60 bên phải.
c)Ta có: \(\left(x-3\right)\left(x-2\right)\left(x+1\right)=60\)
\(\Leftrightarrow\left(x^2-5x+6\right)\left(x+1\right)=60\)
\(\Leftrightarrow x^3+x^2-5x^2-5x+6x+6-60=0\)
\(\Leftrightarrow x^3-4x^2+x-54=0\)
Bạn xem lại đề, nghiệm rất xấu
Tim x, biet:
a, \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
b, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
c, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}\)\(=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
d, \(\dfrac{2}{\left(x-1\right)\left(x-3\right)}+\dfrac{5}{\left(x-3\right)\left(x-8\right)}+\dfrac{12}{\left(x-8\right)\left(x-20\right)}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
Help me!!!!!!!!!!
Can gap lam. Ai lam duoc cau no thi lam nha. Cam on nhieu truoc!!!!!!!!!!!!
a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)
=>x+1=0
hay x=-1
b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)
=>x=15
Tính:
a) \(\dfrac{5^4.20^4}{25^3.4^5}\)
b) \(\left(\dfrac{-10}{3}\right)^5\) .\(\left(\dfrac{-6}{5}\right)^4\)
a) `(5^4 . 20^4)/(25^3 .4^5)`
`=(5^4 . (5.4)^4)/((5^2)^3 .4^5)`
`= (5^4 . 5^4 . 4^4)/(5^6 . 4^5)`
`= (5^2)/4=25/4`
b) `(-10/3)^5 . (-6/5)^4`
`=-10/3 . [(-10/3) . (-6/5)]^4`
`= -10/3 . [ (-5.2 . (-2).3)/(3.5)]^4`
`=-10/3 . 4^4`
`=-2560/3`
A) \(=\dfrac{\left(5.20\right)^4}{\left(25.4\right)^5}=\dfrac{100^4}{100^5}=\dfrac{1}{100}\)
B)=\(\left(\dfrac{-10}{3}\right).\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-6}{5}\right)^4\)
=\(\left(\dfrac{-10}{3}\right).\left(\dfrac{-10}{3}.\dfrac{-6}{5}\right)^4\)
=\(\left(\dfrac{-10}{3}\right).\left(\dfrac{60}{15}\right)^4\)
=\(\left(\dfrac{-10}{3}\right).4^4\)
=\(\left(\dfrac{-10}{3}\right).256\)
=\(\dfrac{-2650}{3}\)
a) \(\dfrac{5^4.20^4}{25^3.4^5}=\dfrac{5^4.4^4.5^4}{5^6.4^5}=\dfrac{5^2}{4}=\dfrac{25}{4}\)
b) \(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{5}\right)^4=\left(\dfrac{-10}{3}\right)^4.\left(\dfrac{-10}{3}\right).\left(\dfrac{-6}{5}\right)^4=\left(\dfrac{-10}{3}.\dfrac{-6}{5}\right)^4.\left(\dfrac{-10}{3}\right)=\left(4\right)^4.\left(\dfrac{-10}{3}\right)=256.\left(\dfrac{-10}{3}\right)=\dfrac{-2560}{3}\)
giải phương trình:
a) \(\left(\dfrac{x}{2}+1\right)^3-\dfrac{x^3}{2}-4=0\)
b) \(\dfrac{3-x}{2007}+1=\dfrac{2-x}{2008}-\dfrac{x}{2009}\)
Lời giải:
a)
PT \(\Leftrightarrow \frac{(x+2)^3}{8}-\frac{x^3+8}{2}=0\)
\(\Leftrightarrow (x+2)^3-4(x^3+8)=0\)
\(\Leftrightarrow (x+2)^3-4(x+2)(x^2-2x+4)=0\)
\(\Leftrightarrow (x+2)[(x+2)^2-4(x^2-2x+4)]=0\)
\(\Leftrightarrow (x+2)(-3x^2+12x-12)=0\)
\(\Leftrightarrow (x+2)(x^2-4x+4)=0\Leftrightarrow (x+2)(x-2)^2=0\Rightarrow x=\pm 2\)
b) Bạn kiểm tra lại xem có sai đề không?
Tìm x; y; z :
a) \(2009-\left|x-2009\right|=x\)
b) \(\left(2x-1\right)^{2008}+\left(y-\dfrac{2}{5}\right)^{2008}+\left|x+y-z\right|=0\)
a: =>|x-2009|=2009-x
=>x-2009<=0
=>x<=2009
b: =>2x-1=0 và y-2/5=0 và x+y-z=0
=>x=1/2 và y=2/5 và z=x+y=1/2+2/5=5/10+4/10=9/10